What will happen if `LiAlH_(4)` is added to an ester `:`

Options

A. Two units of alcohol are obtained
B. One unit of alchol and one unit of acid is obtained
C. Two units of acids are obtained
D. None of these

Answer

To determine what happens when lithium aluminum hydride (LiAlH₄) is added to an ester, we can break down the reaction into several steps.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We start with an ester, which has the general structure RCOOR', where R and R' are hydrocarbon chains or groups.

2. **Understanding LiAlH₄**:
- LiAlH₄ is a strong reducing agent. It can donate hydride ions (H⁻) to other molecules, facilitating the reduction process.

3. **First Step of Reaction**:
- When LiAlH₄ is added to the ester, the hydride ion attacks the carbonyl carbon (C=O) of the ester. This forms a tetrahedral intermediate.
- The double bond between carbon and oxygen shifts, resulting in the formation of an alkoxide ion (RCO⁻OEt) and a lithium ion (Li⁺).

4. **Leaving Group Departure**:
- The alkoxide ion can now expel the leaving group (R'O⁻), which is typically an alcohol (R'OH). This results in the formation of an aldehyde (RCHO).

5. **Second Step of Reaction**:
- The aldehyde formed can undergo further reduction. Another hydride ion from LiAlH₄ attacks the carbonyl carbon of the aldehyde, forming another tetrahedral intermediate.

6. **Final Product Formation**:
- This intermediate then collapses, leading to the formation of a primary alcohol (RCH₂OH) after the workup step, which typically involves protonation (addition of H⁺) to neutralize the alkoxide.

7. **Overall Reaction**:
- The overall result of adding LiAlH₄ to an ester is the formation of two moles of alcohol: one from the ester and one from the aldehyde.

### Final Products:
- The final products of the reaction are two molecules of alcohol (one primary alcohol from the aldehyde and one from the original ester).

### Summary:
When LiAlH₄ is added to an ester, it reduces the ester to yield two molecules of alcohol.

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